descent and essential descent spectrum of linear relations · descent and essential descent...

E extracta mathematicae Vol. 29, Num. 1 - 2, 117 – 139 (2014)

Descent and Essential Descent Spectrumof Linear Relations

Ezzeddine Chafai, Maher Mnif

Departement de Mathematiques, Faculte des Sciences de Sfax, Universite de Sfax,Route de Soukra Km 3.5, B.P. 1171, 3000, Sfax, Tunisia

[email protected], [email protected]

Presented by Manuel Gonzalez Received May 7, 2014

Abstract : In this paper, we study the descent spectrum and the essential descent spectrumof linear relations everywhere defined on Banach spaces. We prove that the correspondingspectra are closed and we obtain that a Banach space X is finite dimensional if and onlyif the descent and the essential descent of every closed linear relation acting in X is finite.We give characterizations of the descent and the essential descent of linear relations and asapplications, some perturbation results are presented.

Key words: Descent, essential descent, spectrum, linear relations.

AMS Subject Class. (2010): 47A06, 47A53, 47A10.

1. Introduction

The notion of descent and essential descent of linear operators was studiedin several articles, for instance, we cite [5],[6], [11], [13] and [22]. In [3],[10] and [18], this concept is extended to the case of linear relations. Manyproperties of descent for the case of linear operators remain be valid in thecontext of linear relations, sometimes under supplementary conditions. M.Burgos, A. Kaidi, M. Mbekhta and M. Oudghiri in [6] and O. Bel Hadj Fredjin [5] studied the descent spectrum and the essential descent spectrum of anoperator acting in Banach spaces. They show that the corresponding spectraare compact subsets of the spectrum, and that for T a bounded operatorin X, σdes(T ) (respectively σe

des(T )) is empty precisely when T is algebraic.Furthermore, they establish that the descent (respectively essential descent)of every operator acting in X is finite if and only if X has finite dimension.The purpose of this paper is to extend the results of the type mentioned aboveto multivalued linear relations in Banach spaces.

To make the paper easily accessible some results from the theory of linearrelations in normed spaces due to R.W. Cross [9] and A. Sandovici, H. de Snoo,

117

118 e. chafai, m. mnif

H. Winkler [18] are recalled in Section 2. In particular, results concerningthe descent of a linear relation are presented. Section 3 is devoted to thestudy of the descent spectrum and the essential descent spectrum of a closedlinear relation T everywhere defined in a Banach space X. We show that thecorresponding spectra are two closed subsets of σ(T ), and that they are emptyprecisely when the essential descent resolvent contains the boundary of thespectrum of T .

Throughout this paper, X will be, unless otherwise stated, an infinitedimensional complex Banach space. A multivalued linear operator in X orsimply a linear relation in X, T : X → X is a mapping from a subspaceD(T ) ⊂ X, called the domain of T , into the collection of nonempty subsetsof X such that T (α1x1 + α2x2) = α1Tx1 + α2Tx2 for all nonzero α1, α2

scalars and x1, x2 ∈ D(T ). We denote the class of linear relations in X byLR(X). If T maps the points of its domain to singletons, then T is said tobe single-valued or simply operator. A linear relation T ∈ LR(X) is uniquelydetermined by its graph, G(T), which is defined by G(T ) := (x, y) ∈ X×X :x ∈ D(T ) and y ∈ Tx. We say that T is closed if its graph is a closedsubspace of X × X, furthermore the class of all closed linear relations in Xwill be denoted by CR(X). The inverse of T is the linear relation T−1 given byG(T−1) := (y, x) : (x, y) ∈ G(T ). The subspace T−1(0), denoted by N(T ),is called the null space of T and we say that T is injective if N(T ) = 0. Therange of T is the subspace R(T ) := T (X) and T is said to be surjective if itsrange coincides with X. We have the following identities

D(T−1) = R(T ), R(T−1) = D(T ), N(T−1) = T (0).

Furthermore, we define the nullity and the defect of T by

α(T ) := dim(N(T )) and β(T ) := codim(R(T )),

respectively.Let M be a subspace of X such that M ∩D(T ) = ∅. Then the restriction

of T to M , denoted by T/M , is given by G(T/M ) := (m, y) ∈ G(T ) : m ∈M ∩ D(T ). For linear relations S and T such that D(T ) ∩ D(S) = ∅, thesum S + T is the linear relation given by G(S + T ) := (x, y + z) : (x, y) ∈G(S), (x, z) ∈ G(T ) and the relation λT , for λ ∈ C, is defined by G(λT ) :=(x, λy) : (x, y) ∈ G(T ). While T − λ stands for T − λI, where I is theidentity operator on X. Let S, T ∈ LR(X), the product ST is defined as therelation G(ST ) := (x, z) : (x, y) ∈ G(T ), (y, z) ∈ G(S) for some y ∈ X.The product of relations is clearly associative. Hence for T ∈ LR(X) and

descent and essential descent spectrum 119

n ∈ Z, Tn is defined as usual with T 0 = I and T 1 = T . It is easily seen that(T−1)n = (Tn)−1, n ∈ Z.

We define the singular chain manifold of T ∈ LR(X) (see [18, (3.3)]) by

Rc(T ) :=

(+∞∪n=0

N(Tn)

)∩

(+∞∪n=0

Tn(0)

).

We say that T has trivial singular chain if Rc(T ) = 0. Let M and N besubspaces of X and X ′ (the dual space of X) respectively. Then M⊥ := x′ ∈X ′ : x′(M) = 0 and N⊤ := x ∈ X : x′x = 0,∀x′ ∈ N. The adjoint T ′ of Tis defined by

G(T ′) := G(−T−1)⊥ ⊂ X ′ ×X ′.

We let QT : X → X/T (0)

be the natural quotient map on X with kernel T (0).

Clearly QTT is a linear operator and that T ∈ LR(X) is closed if and only ifQTT is closed and T (0) is closed in R(T ) (see [9, II.5.3]). For x ∈ X we define∥Tx∥ := ∥QTTx∥ and thus ∥T∥ = sup∥x∥≤1 ∥Tx∥ = ∥QTT∥. We note thatthis quantity is not a true norm since ∥T∥ = 0 does not imply T = 0. T is saidto be continuous if for each open set V in R(T ), T−1(V ) is an open set in D(T )equivalently ∥T∥ < ∞, open if its inverse is continuous equivalently γ(T ) > 0,where γ(T ) := supλ ≥ 0 : λd(x,N(T )) ≤ ∥Tx∥, x ∈ D(T ). Continuouseverywhere defined linear relations are referred to as bounded relations. LetT ∈ CR(X), we say that T is lower semi-Fredholm, if R(T ) is closed andcodimR(T ) is finite.

For T ∈ LR(X), the kernels and the ranges of the iterates Tn, n ∈ N, formtwo increasing and decreasing chains, respectively, i.e the chain of kernels

N(T 0) = 0 ⊂ N(T ) ⊂ N(T 2) ⊂ · · · ,

and the chain of ranges

R(T 0) = X ⊃ R(T ) ⊃ R(T 2) ⊃ · · · .

Furthermore, (see [18, Lemma 3.4]), ifN(T k) = N(T k+1) for some k ∈ N, thenN(Tn) = N(T k) for all nonnegative integers n ≥ k and if R(T k) = R(T k+1)for some k ∈ N, then R(Tn) = R(T k) for all nonnegative integers n ≥ k. Thisstatement leads to define the ascent of T by


a(T ) := infn ∈ N : N(Tn) = N(Tn+1)

,

whenever these minima exist. If no such numbers exist the ascent of T isdefined to be ∞. Likewise, this statement leads to the introduction of thedescent of T by

d(T ) := infn ∈ N : R(Tn) = R(Tn+1)

,

the infimum over the empty set is taken to be ∞. For T ∈ LR(X), we considerthe decreasing sequence βn(T ) := dimR(Tn)/R(Tn+1), n ∈ N (see Lemma 2.3below). We shall say that T has finite essential descent if

de(T ) := infn ∈ N : βn(T ) < ∞

,

where the infimum over the empty set is taken to be infinite, is finite. Clearlyd(T ) = infn ∈ N : βn(T ) = 0 and if T ∈ CR(X) is lower semi-Fredholmthen T has finite essential descent precisely we have de(T ) = 0. Evidently, alinear relation with finite descent has a finite essential descent. In the casewhen de(T ) < ∞ we denote q(T ) := infq ∈ N : βn(T ) = βq(T ), ∀n ≥ q.

Let T ∈ CR(X). The resolvent set of T is the set

ρ(T ) :=λ ∈ C : T − λ is injective and surjective

.

The spectrum of T is the set σ(T ) : C\ρ(T ). Recall (see [9, VI.1.3]), that ρ(T )is an open set and hence σ(T ) is a closed subset of C. The ascent resolvent, thedescent resolvent and the essential descent resolvent sets of a linear relationT ∈ LR(X) are respectively defined by

ρasc(T ) :=λ ∈ C : a(T − λ) < ∞ and R((T − λ)a(T−λ)+1) is closed

,

ρdes(T ) :=λ ∈ C : d(T − λ) < ∞

,

ρedes(T ) :=λ ∈ C : de(T − λ) < ∞

.

The complementary sets σasc(T ) := C\ρasc(T ), σdes(T ) := C\ρdes(T ) andσedes(T ) := C\ρedes(T ) are the ascent spectrum, the descent spectrum and the

essential descent spectrum of T , respectively.


2. Preliminary and auxiliary results

In this section we collect some results of the theory of multivalued linearoperators which are used to prove the main results in Section 3. The proof ofthe next lemma can be found in [9].

Lemma 2.1. ([9, I.3.1]) Let T ∈ LR(X,Y ). Then

(i) TT−1(M) = M ∩R(T ) + T (0), for all M ⊂ Y ,

(ii) T−1T (M) = M ∩D(T ) +N(T ), for all M ⊂ X,

(iii) T (M +N) = T (M) + T (N), for all M ⊂ X and N ⊂ D(T ).

The next lemma is elementary but it is essential to prove Lemma 2.3.

Lemma 2.2. ([18, Lemma 2.1 and Lemma 2.3]) LetM andN be subspacesof a vector space X. Then

(i) dimM/M ∩N = dim(M +N)/N .

(ii) Assume that N ⊂ M then dimX/N = dimX/M + dimM/N.

Let T be a linear relation acting on X. The sequence βn(T ) :=dimR(Tn)/R(Tn+1) play the fundamental role in the definition of the essen-tial descent of T . In the next lemma we show that this sequence is decreasing.

Lemma 2.3. Let T ∈ LR(X). Then

(i) dimR(Tn)/R(Tn+1) ≤ dimR(Tn−1)/R(Tn) for all n ≥ 1.

(ii) If there exists n ∈ N such that dimR(Tn)/R(Tn+1) is finite, thendimR(Tm)/R(Tm+1) is finite for all m ≥ n.

(iii) dimR(Tn)/R(Tn+1) < ∞ if and only if dimR(Tn)/R(Tn+k) < ∞ forall k ≥ 1.

Proof. (i) Let y1, y2, . . . , yk ∈ R(Tn) such that y1, y2, . . . , yk are linearlyindependent in Mn+1 = R(Tn)/R(Tn+1). Then there exist x1, x2, . . . , xk ∈R(Tn−1) such that yi ∈ Txi for all 1 ≤ i ≤ k. Let α1, α2, . . . , αk be scalarssuch that

0 =k∑

i=1

αixi =k∑

i=1

αixi in Mn.


Then∑k

i=1 αixi ∈ R(Tn). It follows that∑k

i=1 αiyi ∈ R(Tn+1), and hence∑ki=1 αiyi = 0. Thus αi = 0 for all 1 ≤ i ≤ k, which implies that x1, x2, . . . xk

are linearly independent in Mn. Now for any k linearly independent vectorsin Mn+1 there exist k linearly independent vectors in Mn. This implies thatdimMn+1 ≤ dimMn.

(ii) By induction, suppose dimR(Tm)/R(Tm+1) < ∞ for some m ≥ n. Then,by (i), dimR(Tm+1)/R(Tm+2) ≤ dimR(Tm)/R(Tm+1) < ∞.

(iii) Assume that dimR(Tn)/R(Tn+1) < ∞ then by Lemma 2.2,

dimR(Tn)/R(Tn+k) =k−1∑i=0

dimR(Tn+i)/R(Tn+i+1) < ∞.

We shall make frequent use of the following result which is the multivaluedversion of the corresponding result for operators. It is used essentially to proveTheorem 3.1.

Lemma 2.4. ([18, Lemma 4.1 and Lemma 4.4]) Let T ∈ LR(X) be every-where defined and let n,m ∈ N. Then

(i) R(Tm)/R(Tm+n) ∼= X/R(Tn) +N(Tm).

(ii) If, moreover, Rc(T ) = 0, then N(Tm+n)/N(Tn) ∼= N(Tm) ∩R(Tn).

There exist closed linear relations S and T such that S + T is not closed.We shall use the following result which gives sufficient conditions for the sumof two closed linear relations to be closed.

Lemma 2.5. ([4, Lemma 14]) Let X be a Banach space and let S, T ∈CR(X) be continuous with S(0) ⊂ T (0) and D(T ) ⊂ D(S). Then S + T isclosed.

The proof of the next result can be found in [18].

Lemma 2.6. ([18, Lemma 7.2]) Let T ∈LR(X). Then N(T −λ)n⊂R(Tm)for all n,m ∈ N and for each λ = 0.

An important role is played by certain root manifolds of a relation T inX. We have the following results which are sometimes useful.


Lemma 2.7. Let T ∈ LR(X). Then

(i) If Rc(T ) = 0, then Rc(T/M ) = 0 for all subspaces M of X.

(ii) Rc(T ) = 0 if and only if Rc(T − λ) = 0, for all λ ∈ C.(iii) ρ(T ) = ∅ if and only if ρ(T − λ) = ∅, for all λ ∈ C.(iv) If ρ(T ) = ∅, then Rc(T ) = 0.

Proof. (i) and (ii) are proved in [20, Lemma 3.1 and Lemma 7.1]. (iii) istrivial and (iv) is a direct consequence of [18, Lemma 6.1] .

It is very well known that for a bounded operator in X, T−1(M) is closedwhenever M is a closed subspace of X. In the following we give conditionsfor T−1(M) to be closed if T is a multivalued linear operator.

Lemma 2.8. Let T ∈ CR(X) be everywhere defined and letM be a closedsubspace of X such that T (0) ⊂ M . Then T−1(M) is closed.

Proof. Since T is closed and everywhere defined then T is bounded(see [9, III.4.2]), and hence QTT is a bounded operator. On the other handQT (M) = (M + T (0))/T (0) = M/T (0) is closed (as M and T (0) are bothclosed, see [4, Lemma 13]). Hence (QTT )

−1QT (M) is closed. But trivially wehave that (QTT )

−1QT (M) = T−1(M+N(QT )) = T−1(M+T (0)) = T−1(M).

The next lemma is used in order to show Lemma 3.8.

Lemma 2.9. Let T ∈ CR(X) be everywhere defined and let M be a closedsubspace of Y such that T (0) ∩M = 0 or T (0) ⊂ M . Suppose M + R(T )and M ∩ R(T ) are closed. If either T (0) or M ∩ R(T ) has finite dimensionthen, R(T ) is closed.

Proof. Write for short N = (M+T (0))∩R(T ) = M∩R(T )+T (0) if T (0)∩M = 0 and N = M ∩R(T ) if T (0) ⊂ M . Clearly N is a closed subspace ofX such that T (0) ⊂ M and hence T−1(N) is closed (by Lemma 2.8). Now wedefine the linear relation

T : (X/T−1(N))⊕ (M/N) → (R(T ) +M)/N

byT (x+m) := y +m : y ∈ Tx.


It is easy to check that T is correctly defined and moreover T is single valued.In fact, for all y ∈ T (0), y = 0 (as T (0) ⊂ N), so that T (0) = 0. Onthe other hand clearly T is surjective and injective. Indeed injective, letx ∈ X and m ∈ M such that T (x + m) = 0 then Tx + m ⊂ N and henceTx ⊂ (M +N) ∩R(T ) = N . It follows that m ∈ N and x ∈ T−1(N), so thatx +m = 0. Since TT−1(N) = N then for all x ∈ X,m ∈ M and y ∈ Tx wehave

∥T (x+m)∥ = ∥y +m∥ = d(y +m,N)

≤ d(y,N) + d(m,N)

= d(y, TT−1(N)) + d(m,N) = infx′∈T−1(N)

d(y, Tx′) + d(m,N).

Which implies that

∥T (x+m)∥ ≤ infx′∈T−1(N)

d(Tx, Tx′) + d(m,N)

= infx′∈T−1(N)

∥Tx− Tx′∥+ d(m,N)

≤ ∥T∥ infx′∈T−1(N)

∥x− x′∥+ d(m,N)

= ∥T∥d(x, T−1(N)) + d(m,N)

≤ (1 + ∥T∥)(d(x, T−1(N)) + d(m,N)

)= (1 + ∥T∥)∥x+m∥.

Thus T is bounded and since T is bijective then, by the Mapping Theoremof linear operators, T (X/(T−1(N))) is closed. Now let P : R(T ) + M →(R(T ) +M)/N be the canonical projection. Then R(T ) = P−1(R(T )/N) =P−1(T (X/T−1(N)). Thus R(T ) is closed.

Lemma 2.10. Let X,Y be two Banach spaces and T ∈ CR(X). ThenD(T ) endowed with the norm | · | defined by

|x| := ∥x∥+ inf∥y∥ : y ∈ Tx = ∥x∥+ ∥Tx∥,

is a Banach space.

Proof. Since T is closed then QT is closed (see [9, Proposition II.5.3]),which implies that G(QT ) is closed. On the other hand, T (0) is closed (asT is closed) so that Y/T (0) is a Banach space. It follows that G(QT ) iscomplete. Now, clearly (D(T ), | · |) and G(QT ) are isomorphic. This provethat (D(T ), | · |) is complete.


3. Characterization of the descent andthe essential descent spectrum

In this section we give characterization and some properties of the descentspectrum and the essential descent spectrum of linear relations everywhere de-fined in Banach spaces. We investigate that the corresponding spectrums areclosed and that the descent (respectively essential descent) of every multival-ued linear operator acting in X is finite if and only if X has finite dimension.For this end, we first prove some technical lemmas.

Definition 3.1. Let T ∈ LR(X), let k and mi, 1 ≤ i ≤ n be somepositive integers, and let λi ∈ C, 1 ≤ i ≤ k be some distinct constants. LetP (X) =

∏ki=1(X − λi)

mi . Then the polynomial P in T is the linear relation

P (T ) :=k∏

i=1

(T − λi)mi .

The behaviour of the domain, the range, the null space and the multivaluedpart of P (T ) is described in the following lemma which is due to Sandovici[19].

Lemma 3.1. ([19, 3.2, 3.3, 3.4, 3.5 and 3.6]) Let T be a linear relationin a vector space X, let n ∈ N, λi ∈ K,mi ∈ N, 1 ≤ i ≤ k. Assume that λi,1 ≤ i ≤ k are distinct and let P (T ) be as in Definition 3.1. Then

(i) D(P (T )) = D(T∑k

i=1 mi).

(ii) R(P (T )) =

k∩i=1

R(T − λi)mi .

(iii) N(P (T )) =

k∑i=1

N(T − λi)mi .

(iv) (T − λ)k(0) = T k(0), if λ ∈ K.

(v) P (T )(0) = T∑k

i=1 mi(0).

As consequence of the above lemma we have the next result which will beused in the sequel.


Lemma 3.2. Let T be a linear relation on X everywhere defined. Let

P (X) =k∏

i=1

(X − λi) = Xk +k−1∑j=0

ajXj

and

Q(X) =k∏

i=1

(X − µi) = Xk +k−1∑j=0

bjXj .

Then

(i) P (T ) = T k +

k−1∑j=0

ajTj .

(ii) P (T )−Q(T ) = (P −Q)(T ) + T k − T k.

(iii) P (T )Q(T ) = (PQ)(T ).

The following purely algebraic lemma helps to read Definition 3.2 below.There exhibits some useful connections between the kernels and the ranges ofthe iterates Tn of a linear relation T on X.

Lemma 3.3. ([14, Lemma 3.7]) Let T ∈ LR(X). Then the following state-ments are equivalent

(i) N(T ) ⊂ R(Tn) for each n ∈ N.

(ii) N(Tm) ⊂ R(T ) for each m ∈ N.

(iii) N(Tm) ⊂ R(Tn) for each m ∈ N and n ∈ N.

Definition 3.2. We say that a linear relation T ∈ LR(X) is regular ifR(T ) is closed and T verifies one of the equivalent conditions of Lemma 3.3.

Trivial examples of regular linear relations are surjective multivalued op-erators as well as injective multivalued operators with closed range.

Lemma 3.4. Let T ∈ CR(X) be regular and everywhere defined withfinite codimensional range. Then

codimR(Tn) = n codimR(T )

for all positive integers n.


Proof. Let n ≥ 1. Since T is regular then, X/R(T ) = X/R(T ) +N(Tn−1)and according to Lemma 2.4(i), it follows that X/R(T ) ∼= R(Tn−1)/R(Tn).On the other hand, using Lemma 2.2, we obtain that dim[(X/R(Tn−1)) ×(R(Tn−1)/R(Tn))] = dim(X/R(Tn)), which implies that

codimR(Tn) = dim(X/R(Tn)) = dim[(X/R(Tn−1))× (R(Tn−1)/R(Tn))]

= dim(X/R(Tn−1)) + dim(R(Tn−1)/R(Tn))

= codimR(Tn−1) + codimR(T )

Thus, a successive repetition of this argument leads to codimR(Tn) =n codimR(T ).

As consequence of [8, Theorem 3.1] we have that

Lemma 3.5. Let T ∈ LR(X) be everywhere defined and with trivial sin-gular chain. Then

ind(Tn) = n ind(T ) (3.1)

The next lemma is elementary but essential to prove Theorem 3.1.

Lemma 3.6. Let T ∈ LR(X) be such that de(T ) is finite. Then

N(T ) ∩R(T q) = N(T ) ∩R(T q+n)

for some q ∈ N and for all n ∈ N.

Proof. It is very well known (see [16, Lemma 22.2]) that if U, V and Ware subspaces of a Banach space X such that U ⊂ W then,

(U + V ) ∩W = U + (V ∩W ). (3.2)

Let q = q(T ), and let T : R(T q)/R(T q+1) → R(T q+1)/R(T q+2) be the quo-tient map defined by T x := y : y ∈ Tx. Then T0 = 0 since T (0) ⊂ R(T q+2).It follows that T is single valued and hence T x = y for any (then for all)y ∈ Tx. According to (3.2), Lemma 2.2 and Lemma 2.1, we obtain that

dimN(T ) = dim[(T−1(R(T q+2)) ∩R(T q))/R(T q+1)]

= dim[(N(T ) +R(T q+1)) ∩R(T q)/R(T q+1)]

= dim[((N(T ) ∩R(T q)) +R(T q+1))/R(T q+1)]

= dim[(N(T ) ∩R(T q))/(N(T ) ∩R(T q+1))].


On the other hand it is easy to see that T is surjective. Furthermore, sincedimR(T q)/R(T q+1) = dimR(T q+1/R(T q+2), then T is injective, which im-plies that dimN(T ) = 0. Therefore, N(T ) ∩R(T q) = N(T ) ∩R(T q+1).

The next lemma investigates the stability of certain Fredholm type prop-erties of a linear relation under small perturbation. It is used to prove Theo-rem 3.1.

Lemma 3.7. ([2, Theorem 23, Theorem 25 and Theorem 27]) Let T ∈CR(X) be lower semi-Fredholm and regular. Then there exists δ > 0 suchthat T − λ is both lower semi-Fredholm and regular for all |λ| < δ. Moreoverβ(T − λ) = β(T ) for 0 < |λ| < δ.

In [6], M. Burgos, A. Kaidi, M. Mbekhta and M. Oudghiri prove thatρdes(T ) is an open set for every bounded operator in X. In the following weshow that both ρdes(T ) and ρedes(T ) are open for each T ∈ CR(X) everywheredefined, and if moreover T is regular then σdes\σe

des(T ) is also open.

Theorem 3.1. Let T ∈ CR(X) be everywhere defined for which de(T ) isfinite and suppose ρ(T ) = ∅. Then there exists δ > 0 such that for 0 < |λ| < δand q = q(T ), we have the following assertions:

(i) T − λ is regular.

(ii) codimR(T − λ)n = n codimR(T ) = n dim(R(T q)/R(T q+1)) for alln ∈ N.

(iii) dimN(T − λ)n = n dim(N(T q+1)/N(T q)) for all n ∈ N.

Proof. Since T is closed and ρ(T ) = ∅ then T q is closed [10, Lemma 3.1].It follows that T−q is closed and hence, from Lemma 2.10, R(T q) = D(T−q)equipped with the norm | · | defined by

|y| := ∥y∥+ inf∥x∥ : x ∈ X and y ∈ T qx

= ∥y∥+ ∥T−qy∥,

is a Banach space. Let T0 := T/R(T q). Then T0 is closed and lower semi-Fredholm and hence by [1, Corollary 1.8] R(T0) is closed. Indeed T0 is closed,

let (xn, yn)n be a sequence of G(T0) converging to (x, y). Then xn|·|→x and

yn|·|→ y so that x, y ∈ R(T q). Moreover xn

∥·∥→x and yn∥·∥→ y which implies

that (x, y) ∈ G(T ). Thus (x, y) ∈ G(T0). Further, since R(T0) = R(T q+1) isof finite codimension in R(T q) then T0 is lower semi-Fredholm. According toLemma 3.6, N(T0) = N(T )∩R(T q) = N(T )∩R(T q+n) ⊂ R(T q+n) = R(Tn

0 ).


Thus T0 is regular. On the other hand let α ∈ ρ(T ) then 0 = N(T − α) ⊃N(T − α) ∩R(T q) = N(T0 − α) and R(T0 − α) = (T − α)(R(T q)) = T q[(T −α)(X)] = T q(X) = R(T q) (as T q and T − α commute). This implies thatα ∈ ρ(T0) and thus ρ(T0) = ∅. Furthermore, by Lemma 3.7 there exists δ > 0such that T0 − λ is regular, lower semi-Fredholm and β(T0 − λ) = β(T0), forall 0 < |λ| < δ. Now, for n ≥ 1 and λ = 0, we consider the polynomialsP, Q defined by P (z) = (z − λ)n and Q(z) = zq, for all z ∈ C. Clearly thatP and Q have no common divisors, then there exist two polynomials u andv such that 1 = P (z)u(z) + Q(z)v(z) for all z ∈ C. Hence by Lemma 3.2,I + Tm − Tm = (T − λ)nu(T ) + T qv(T ) for some m ≥ max(n, q) and thusX = R(T − λ)n + R(T q). Since T0 is closed then by Lemma 2.5, T0 − λ isclosed. It follows by Lemma 3.4, Lemma 2.2 and Lemma 3.7 that

codimR(T − λ)n = dimX/R(T − λ)n

= dim[(R(T q) +R(T − λ)n)/R(T − λ)n]

= dim[R(T q)/R(T q) ∩R(T − λ)n]

= codimR(T0 − λ)n

= n codimR(T0 − λ)

= n codimR(T0)

= n dimR(T q)/R(T q+1).

This implies that T − λ is semi-Fredholm. Moreover, by Lemma 2.6, N(T −λ) = N(T − λ) ∩ R(T q) = N(T0 − λ) ⊆ R(T0 − λ)n ⊆ R(T − λ)n, whichimplies that T − λ is regular. For the statement (iii), since ρ(T ) = ∅ thenρ(T0) = ∅ and hence ρ(T0 − λ) = ∅, which implies that Rc(T0 − λ) = 0 (seeLemma 2.7). Therefore, using (3.1), Lemma 3.7 and Lemma 3.4, we have that

dimN(T − λ)n = dimN(T − λ)n ∩R(T q)

= dimN(T0 − λ)n

= ind(T0 − λ)n + codimR(T0 − λ)n

= n[ind(T0 − λ) + codimR(T0 − λ)]

= n[ind(T0) + codimR(T0)]

= n dimN(T0) = n dimN(T ) ∩R(T q)

Since, by Lemma 2.4, N(T q+1)/N(T q) ∼= N(T ) ∩R(T q) then

dimN(T − λ)n = n dim(N(T ) ∩R(T q)) = n dim(N(T q+1)/N(T q)).


As a direct consequence of Theorem 3.1 we obtain the following result forclosed multivalued linear operators everywhere defined and with finite ascent.

Corollary 3.1. Let T ∈ CR(X) be everywhere defined with finite de-scent. Suppose ρ(T ) = ∅. Then there exists δ > 0 such that for all 0 < |λ| < δ,

(i) T − λ is surjective.

(ii) dimN(T − λ) = dimN(T q+1)/N(T q) = dim(N(T ) ∩ R(T q)) for someq ∈ N.

(iii) If, moreover, a(T ) < ∞, then T − λ is bijective.

Corollary 3.2. Let T ∈ CR(X) be everywhere defined with ρ(T ) = ∅.Let λ0 ∈ σ(T ) such that T − λ0 has finite ascent and descent. Then λ0 is anisolated point of σ(T ) and it is in the boundary of the spectrum of T .

Proof. Let λ0 ∈ σ(T ) such that T − λ0 has finite ascent and descent.Clearly by Lemma 2.5 and Lemma 2.7 that T −λ0 is closed and ρ(T −λ0) = ∅.Hence there exists by Corollary 3.1, δ > 0 such that T − µ is injective andsurjective for all 0 < |µ − λ0| < δ. This implies that D(λ0, δ)\λ0 ⊂ ρ(T ).Thus λ0 is isolated and it is in the boundary of the spectrum of T .

Also as a consequence of Theorem 3.1, we have

Corollary 3.3. Let T ∈ CR(X) be everywhere defined and supposeρ(T ) = ∅. Then ρdes(T ) and ρedes(T ) are open subsets and hence σdes(T ) andσedes(T ) are closed subsets of σ(T ). Moreover, if T is regular then

σdes(T )\σedes(T ) is an open set.

Proof. The openness of ρedes(T ) and ρdes(T ) follows directly from Theo-rem 3.1 and Corollary 3.1, respectively. Now let λ ∈ σdes(T )\σe

des(T ) and letq := q(T − λ). We see easily, by Lemmas 2.5 and 2.7, that T − λ is closedand ρ(T − λ) = ∅. Therefore, Theorem 3.1 ensures that there exists an openneighborhood U of λ such that U ∩ σe

des(T ) = ∅ and codimR(T − α)n =n dim(R(T − λ)q/R(T − λ)q+1), for all α ∈ U and n ∈ N. Since T − λ hasinfinite descent, dim(R(T − λ)q/R(T − λ)q+1) is nonzero, and consequently(codimR(T − α)n)n is a strictly increasing sequence for each α ∈ U . ThusU ⊂ σdes(T ), as desired.

For T ∈ LR(X) we define E(T ) := ρdes(T ) ∩ ρasc(T )\ρ(T ) = ρdes(T ) ∩ρasc(T ) ∩ σ(T ). The descent spectrum, and therefore the essential descent


spectrum, of a linear relation can be empty. In the next theorem we showthat this occurs precisely when ∂σ(T ) ⊆ ρedes(T ).

Theorem 3.2. Let T ∈ CR(X) be everywhere defined with ρ(T ) = ∅.Then

ρedes(T ) ∩ ∂σ(T ) = ρdes(T ) ∩ ∂σ(T ) = E(T ). (3.3)

Moreover, the following assertions are equivalent:

(i) σdes(T ) = ∅.(ii) σe

des(T ) = ∅.(iii) ∂σ(T ) ⊆ ρdes(T ).

(iv) ∂σ(T ) ⊆ ρedes(T ).

The proof of this theorem requires the following lemma.

Lemma 3.8. Let T ∈ CR(X) be everywhere defined such that ρ(T ) = ∅.Suppose a(T ) and d(T ) are both finite. Then R(Tn) is closed, for all n ≥ a(T ).

Proof. First observe, since T is closed and ρ(T ) = ∅ then Tn is closed forall n ∈ N (by [10, Lemma 3.1 ]) and hence N(Tn) is closed. On the other handD(T ) = X leads to D(Tn) = X and since T has finite ascent and descent, itfollows by ([18, Theorem 5.7]) that a(T ) = d(T ). Let a := a(T ), d := d(T )and fix n ≥ a. Lemma 5.5 of [18] leads to N(Tn) ∩R(Tn) = 0. Now, sinceT is everywhere defined, Lemma 5.6 of [18] ensures that N(Tn)∩R(Tn) = X.According to Lemma 2.9, it follows that R(Tn) is closed.

Proof of Theorem 3.2. According to Corollary 3.2, the inclusions ρedes(T )∩∂σ(T ) ⊃ ρdes(T )∩∂σ(T ) ⊃ E(T ) are trivial. Now, let λ in the boundary of thespectrum of T such that T−λ has finite essential descent. By Lemma 2.5, T−λis closed and Lemma 2.7 leads to ρ(T − λ) = ∅. Then by Theorem 3.1, thereexists a punctured neighborhood V of λ such that dimN(T−α) = dim(N(T−λ)p)/(N(T−λ)q) and codimR(T−α) = dim(R(T−λ)q/R(T−λ)q+1) for someq ∈ N and for all α ∈ V . Moreover T − α is closed for all α (see Lemma 2.5)and, since λ ∈ ∂σ(T ) and ρ(T ) = ∅, then there exists α0 ∈ V \σ(T ) = ∅.Therefore

0 = dimN(T − α0) = codimR(T − α0)

= dim(N(T − λ)q+1)/N((T − λ)q)

= dim(R((T − λ)q)/R((T − λ)q+1)).


This means that T − λ is of finite ascent and descent. Lemma 3.8 leads toR(T a(T )+1) is closed and therefore λ ∈ E(T ).

All the desired implications (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are clear from (3.3).For (iv) ⇒ (i), assume that ∂σ(T ) ⊆ ρedes(T ) then ∂σ(T ) = E(T ). Accord-ing to Lemma 3.2, it follows that all points of ∂σ(T ) are isolated and hence∂σ(T ) = σ(T ). By (3.3) we conclude that σ(T ) ⊂ ρdes(T ) ∩ ρasc(T ). Whichmeans that C = σ(T ) ∪ ρ(T ) ⊂ ρdes(T ) ∩ ρasc(T ). Hence ρdes(T ) = C, whichmeans that σdes(T ) = ∅.

Corollary 3.4. LetX be a Banach space and letK(X) := T ∈ CR(X) :D(T ) = X, and ρ(T ) = ∅. The following assertions are equivalent:

(i) X is finite dimensional.

(ii) T is of finite descent for every T ∈ K(X).

(iii) T is of finite essential descent for every T ∈ K(X).

(iv) σdes(T ) is empty for every T ∈ K(X).

(v) σedes(T ) is empty for every T ∈ K(X).

Proof. The implications (i)⇒ (ii)⇒ (iii) and (ii)⇒ (iv)⇒ (v) are obvious.Using Lemma 2.7 and Lemma 2.5, it is easy to see that T ∈ K(X) if and onlyif T −λ ∈ K(X) for all λ ∈ C. It follows by Theorem 3.2, that (iii) entails (ii).For (v) ⇒ (i), suppose σe

des(T ) is empty for every T ∈ K(X). Then σedes(T )

is empty for every T ∈ L(X) (the space of all bounded operators on X). Itfollows by [6, Corollary 1.10)] that X is finite dimensional.

Theorem 3.3. Let T ∈ CR(X) be everywhere defined such that ρ(T ) = ∅and let Ω be a connected component of ρedes(T ). Suppose ρ(T ) = ∅. Then

Ω ⊂ σ(T ) or Ω\E(T ) ⊆ ρ(T ).

Proof. Let Ωr :=λ ∈ Ω : T−λ is both regular and lower semi-Fredholm.According to Theorem 3.1, Ω\Ωr is at most countable, and hence Ωr isconnected. Suppose Ω ∩ ρ(T ) is non-empty, then so is Ωr ∩ ρ(T ). Using[4, Corollary 17] and Lemma 3.7 we obtain that codimR(T − λ) = 0 and bythe continuity of the index (see [4, Theorem 15]), we get that N(T−λ) = 0 forall λ ∈ Ωr. Thus Ωr ⊆ ρ(T ). Consequently, Ω\Ωr consists of isolated pointsof the spectrum with finite essential descent. That is,

Ω\Ωr ⊆ ρedes(T ) ∩ ∂σ(T ) = E(T ).

Therefore, Ω\E(T ) ⊆ Ωr ⊆ ρ(T ).


Corollary 3.5. Let T ∈ CR(X) be everywhere defined such that ρ(T ) =∅. The following assertions are equivalent:

(i) σ(T ) is at most countable.

(ii) σdes(T ) is at most countable.

(iii) σedes(T ) is at most countable.

In this case, σedes(T ) = σdes(T ) and σ(T ) = σdes(T ) ∪ E(T ).

Proof. Since the implications (i) ⇒ (ii) ⇒ (iii) are obvious, to finish theproof it suffices to show that (iii) ⇒ (i). Suppose σe

des(T ) is at most countable,then ρedes(T ) is connected. Since ρ(T ) = ∅ it follows, by Theorem 3.3, thatρedes(T )\E(T ) ⊆ ρ(T ). Which implies that σ(T ) = σe

des(T ) ∪ E(T ) is atmost countable. Finally, for the last assertion, assume that σ(T ) is at mostcountable. Then by Corollary 3.3, we obtain that σdes(T )\σe

des(T ) is at mostcountable and open. Therefore it is empty. Thus σe

des(T ) = σdes(T ).

Remark 3.1. An immediate consequence of the previous corollary that ifT ∈ CR(X) be everywhere defined with ρ(T ) = ∅, then σ(T )\0 ⊂ E(T ) ifand only if σdes(T ) ⊂ 0 if and only if σe

des(T ) ⊂ 0.

4. Descent, essential descent spectrum and perturbations

We start this section by the next lemma which are used in the sequel.

Lemma 4.1. Let T ∈ LR(Y,Z) and S,R ∈ LR(X). If T (0) ⊂ N(S) orT (0) ⊂ N(R) then

(R+ S)T = RT + ST.

Proof. Suppose T (0) ⊂ N(R), the case T (0) ⊂ N(S) is similar. Theinclusion (R + S)T ⊂ RT + ST is proved in [9, I.4.2 (d)]. For the reverseinclusion, let x ∈ D(RT +ST ) = D(RT )∩D(ST ) = T−1(D(R))∩T−1(D(S)).Then there exist y1 ∈ Tx∩D(R) and y2 ∈ Tx∩D(S), so that y1−y2 ∈ T (0) ⊂N(R) ⊂ D(R). Hence y2 = (y2 − y1) + y1 ∈ D(R) ∩ D(S) = D(R + S). Itfollows that Tx∩D(R+S) = ∅. Thus x ∈ T−1(D(R+S)) = D(R+S)T andhence

D(RT + ST ) ⊂ D((R+ S)T ). (4.1)


Now we show that (RT + ST )x ⊂ (R + S)Tx, for x ∈ D(RT + ST ). Lety ∈ (RT + ST )x = RTx + STx. Then y ∈ Ry1 + Sy2 for some y1, y2 ∈ Tx.Hence y1 − y2 ∈ T (0) ⊂ N(R). It follows that y ∈ R(y1 − y2 + y2) + S(y2) =R(y1 − y2) + R(y2) + S(y2) = R(0) + Ry2 + Sy2 = Ry2 + Sy2 = (R + S)y2.Hence y ∈ (R+ S)Tx. Thus

(RT + ST )x ⊂ (R+ S)Tx (4.2)

(4.1) and (4.2) implies that RT + ST ⊂ (R + S)T . The reverse inclusion isproved in [9, I.4.2].

Lemma 4.2. Let T ∈ CR(X) be everywhere defined and F be a boundedoperator such that TF = TF . Suppose T (0) ⊂ N(T ) or T (0) ⊂ N(F ). Then,for every n ∈ N,

(i) (T + F )n =

n∑i=0

(n

i

)Tn−iF i, for all n ∈ N,

(ii) Tn − Fn = (

n−1∑i=0

(−1)iTn−1−iF i)(T − F ).

Proof. (i) For n = 0 and n = 1 the result is trivial. Let n ≥ 1 and suppose(T + F )n =

∑ni=0(

ni )T

n−iF i. Using [9, I.4.2 (d) and (e)] and Lemma 4.1, itfollows that

(T + F )n+1 = (T + F )n(T + F ) = (T + F )nT + (T + F )nF

=

( n∑i=0

(n

i

)Tn−iF i

)T +

( n∑i=0

(n

i

)Tn−iF i

)F

=n∑

i=0

(n

i

)Tn−i+1F i +

n∑i=0

(n

i

)Tn−iF i+1

=

n∑i=0

(n

i

)Tn−i+1F i +

n+1∑i=1

(n

i

)Tn−i+1F i

= Tn+1 +n∑

i=1

(n+ 1

i

)Tn−i+1F i + Fn+1

=n∑

i=0

(n+ 1

i

)Tn+1−iF i.


(ii) First observe that Tn +T kF j −T kF j = Tn for all 0 ≤ k ≤ n and j ∈ N.It follows that( n−1∑

i=0

Tn−1−iF i

)(T − F )

=(Tn−1 + Tn−2F + · · ·+ Fn−1

)T −

(Tn−1 + Tn−2F + · · ·+ Fn−1

)F

=(Tn + Tn−1F + · · ·+ TFn−1

)−(Tn−1F + Tn−2F 2 + · · ·+ Fn

)= Tn − Fn +

(Tn−1F − Tn−1F + Tn−2F 2 − Tn−2F 2+

· · ·+ TFn−1 − TFn−1)

= Tn − Fn.

Lemma 4.3. Let F be an operator and T ∈ LR(X) be everywhere definedsuch that T (0) ⊂ N(T ). Suppose that F commutes with T . Then

dimR(Tn+k−1)/R(T + F )n ∩R(Tn+k−1) ≤ dimR(F k), for all n, k ≥ 1.

Proof. Let y1, y2, . . . , ym be in R(Tn+k−1) such that y1, y2, . . . , ym are lin-early independent inR(Tn+k−1)/R(Tn+k−1)∩R(T+F )n. There exist elementsx1, x2, . . . , xm such that yi ∈ Tn+k−1xi for 1 ≤ i ≤ m. Since F is single-valuedand everywhere defined then (T + F ) and (−F ) commute. We replace T byT + F and F by −F in Lemma 4.2, it follows that yi ∈ (T + F )nzi + F kti,for suitable zi and ti. Suppose m > dimR(F k), then there exist constantsα1, α2, . . . , αm, not all zero such that

∑mi=1 αiF

kti = 0. Thus

m∑i=1

αiyi ∈m∑i=1

(T + F )nzi ⊂ R(T + F )n.

Furthermore, since the αi are not all zero, y1, y2, . . . , ym are not linearly in-dependent in R(Tn+k−1)/R(Tn+k−1) ∩ R(T + F )n, a contradiction. Thusm ≤ dimR(F k).

Now, we are in the position to give the main theorem of this section.

Theorem 4.1. Let F be a bounded operator acting in X such that F k isof finite rank for some nonnegative integer k and let T ∈ LR(X) be everywheredefined which commutes with F . Suppose T (0) ⊂ N(T ) then

(i) de(T ) is finite if and only if de(T + F ) is finite.


(ii) d(T ) is finite if and only if d(T + F ) is finite.

Proof. (i) Since F k has finite-dimensional rank then, by Lemma 4.3,

dimR(Tn+k−1)/R(T + F )n ∩R(Tn+k−1) ≤ dimR(F k) < ∞, (4.3)

for all positive integer n. Let d := de(T ) < ∞, then by Lemma 2.3,

dimR(T d)/R(Tn+k−1) < ∞, for all n ≥ d. (4.4)

A combination of (4.3) and (4.4) leads to, dimR(T d)/R(T +F )n∩R(Tn+k−1)is finite for n ≥ d. On the other hand R(T +F )n∩R(Tn+k−1) ⊂ R(T +F )n∩R(T d) ⊂ R(T d). Which implies that dimR(T d)/R(T + F )n ∩ R(T d) is finiteand as dimR(F k) is finite we obtain that,

dim((R(T d) +R(F k))/R(T + F )n ∩R(T d)

)< ∞, for all n ≥ d. (4.5)

If we replace T by T + F and F by −F in (4.3), we get that

dimR(T + F )n+k−1/R(Tn) ∩R(T + F )n+k−1 < ∞.

It follows that

dimR(T + F )n+k−1/R(T d) ∩R(T + F )n+k−1 < ∞, for all n ≥ d. (4.6)

Combining (4.5) and (4.6) we obtain that

dim((R(T d) +R(F k))/R(T + F )n

)< ∞, for all n ≥ d+ k.

Therefore

dim[R(T + F )n/R(T + F )n+1] = dim[(R(T d) +R(F k))/R(T + F )n+1]

− dim[(R(T d) +R(F k))/R(T + F )n] < ∞,

for all n ≥ d+ k. Thus de(T +F ) < ∞. For the reverse implication it sufficesto replace T by T + F and F by −F .

(ii) Let d = d(T ) < ∞, and for n ≥ d, let

an(T ) : = dimR(Tn+k−1)/R(T + F )n ∩R(Tn+k−1)

= dimR(T d)/R(T + F )n ∩R(T d),

bn(T ) : = dimR(T + F )n+k−1/R(T + F )n+k−1 ∩R(Tn)

= dimR(T + F )n+k−1/R(T + F )n+k−1 ∩R(T d).


By Lemma 4.3, an(T ) ≤ dimR(F k) < ∞. Furthermore clearly (an(T ))n isan increasing sequence and hence there exists an integer p such that an(T ) =ap(T ), for n ≥ p. It follows that

R(T + F )p ∩R(T d) = R(T + F )i ∩R(T d), for i ≥ p. (4.7)

By interchanging T by T + F and F by −F in Lemma 4.3, we obtain thatbn(T ) ≤ dimR(F k) < ∞. Furthermore it is easy to see that (bn(T ))n is anincreasing sequence and consequently there exists q ≥ p such that bn(T ) =bq(T ) for n ≥ q. Now using (4.7) it follows that, for n ≥ q ≥ p,

dimR(Tn+k−1)/R(T + F )p ∩R(T d)

= dimR(Tn+k−1)/R(T + F )n+k−1 ∩R(T d)

= dimR(T q+k−1)/R(T + F )q+k−1 ∩R(T d)

= dimR(T q+k−1)/R(T + F )p ∩R(T d).

This implies thatR(T+F )n+k−1 = R(T+F )q+k−1, for n ≥ q. Thus d(T+F ) <q + k < ∞.

Corollary 4.1. Let F ∈ LR(X) be single valued and bounded and letKF :=

T ∈ LR(X) : D(T ) = X, TF = FT and T (0) ⊂ N(T )

. Then the

following assertions are equivalent:

(i) F k is of finite rank for some integer k.

(ii) σdes(T + F ) = σdes(T ) for every T ∈ KF .

(iii) σedes(T + F ) = σe

des(T ) for every T ∈ KF .

The proof of this corollary requires the following lemma.

Lemma 4.4. ([5, Theorem 3.1] and [6, Theorem 3.1]) Let F ∈ LR(X) bea bounded operator. Then the following assertions are equivalent:

(i) There exists a positive integer k for which F k is of finite rank.

(ii) σdes(T +F ) = σdes(T ) for every bounded operator T ∈ LR(X) commut-ing with F.

(iii) σedes(T +F ) = σe

des(T ) for every bounded operator T ∈ LR(X) commut-ing with F.


Proof of Corollary 4.1. (i) ⇒ (ii) and (i) ⇒ (iii) are an immediate conse-quences of Theorem 4.1. For the implications (ii) ⇒ (i) it suffices to see thatσdes(T+F ) = σdes(T ) for every T ∈ KF implies that σdes(T+F ) = σdes(T ) forevery bounded operator T ∈ LR(X) commuting with F and using Lemma 4.4it follows that (i) holds. Similarly Lemma 4.4 leads to (iii) ⇒ (i). This com-pletes the proof.

Also as an immediate consequence of Theorem 4.1 we have that

Corollary 4.2. Let T ∈ LR(X) be everywhere defined. Then

σedes(T ) ⊂

∩F∈FT (X)

σdes(T + F ),

where FT (X) denotes the set of all bounded finite-rank operators F on Xcommuting with T and such that T (0) ⊂ N(T ).

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